The generic reaction 2A rightarrow B has the following rate laws: forward reacti

Question

The generic reaction 2A rightarrow B has the following rate laws: forward reaction: rate = kf [A]2 reverse reaction: rate = kr[B] where kf is the rate constant for the forward reaction and kr is the rate constant for the reverse reaction. At equilibrium, the two rates are equal and so kf [A]2 = kr[B]. The equilibrium constant for a reaction is related by the law of mass action to the rate constants for the forward and reverse reactions: Kc = [B] / [A]2 = kf/kr Nitrosyl bromide, NOBr, is formed in the reaction of nitric oxide, NO, with bromine, Br2: 2NO(g) + Br2(g) 2NOBr(g) The reaction rapidly establishes equilibrium when the reactants are mixed. At a certain temperature the concentration of NO was 0.400 M and that of Br2 was 0.255 M. At equilibrium the concentration of NOBr was found to be 0.250 M. What is the value of Kc at this temperature? Express your answer numerically.

Explanation / Answer

The person above me set up the problem correctly, the mathmight be off. it should be 0.252 / (0.152*0.13) =21.36752137

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